Дано: m=1,2m0

Найти: l0 / l -

Решение:

$m=\frac{m_0}{\sqrt{1-(\frac{v}{c})^2}}\\ 1,2m_0=\frac{m_0}{\sqrt{1-(\frac{v}{c})^2}}\\ \sqrt{1-(\frac{v}{c})^2} = \frac{m_0}{1,2m_0}\\ \sqrt{1-(\frac{v}{c})^2} = \frac{5}{6}\\ 1-(\frac{v}{c})^2=\frac{25}{36}\\ (\frac{v}{c})^2=\frac{9}{36}\\ v^2=\frac{9c^2}{36}\\ v=\sqrt{\frac{9c^2}{36}}=\frac{3c}{6}=\frac{c}{2}\\ \\ l=l_0\sqrt{1-(\frac{v}{c})^2}=l_0\sqrt{1-\frac{1}{4}}=\frac{\sqrt3}{2}l_0=0,8l_0\\ \frac{l_0}{l}=\frac{l_0}{0,8l_0}=1,25$

Ответ: в 1,25 раза.